hdu 1016 Prime Ring Problem
2014-08-11 15:55
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参考博文:/article/4984202.html
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26786 Accepted Submission(s): 11950
Problem Description
A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.
Note: the number of first circle should always be 1.
Input
n (0 < n < 20).
Output
The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.
You are to write a program that completes above process.
Print a blank line after each case.
Sample Input
Sample Output
Source
Asia 1996, Shanghai (Mainland China)
这题的思路就写在代码里了。刚看到题时,要说什么想法都没有还是有一点儿的,但又不知从何入手,所以参考了前辈的代码,相应的原文链接就在文首,代码也几乎是复制版本(先仔细推敲前辈的代码,开启记忆功能,然后自己一点儿点儿的近乎默写的形式敲出了几乎全部代码,测试时有问题又回头比照了原文。) 再注释一遍,权当复习,以后回头望时还有个痕迹~
参考代码:
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 26786 Accepted Submission(s): 11950
Problem Description
A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.
Note: the number of first circle should always be 1.
Input
n (0 < n < 20).
Output
The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.
You are to write a program that completes above process.
Print a blank line after each case.
Sample Input
6 8
Sample Output
Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2
Source
Asia 1996, Shanghai (Mainland China)
这题的思路就写在代码里了。刚看到题时,要说什么想法都没有还是有一点儿的,但又不知从何入手,所以参考了前辈的代码,相应的原文链接就在文首,代码也几乎是复制版本(先仔细推敲前辈的代码,开启记忆功能,然后自己一点儿点儿的近乎默写的形式敲出了几乎全部代码,测试时有问题又回头比照了原文。) 再注释一遍,权当复习,以后回头望时还有个痕迹~
参考代码:
#include<iostream> #include<cstdio> using namespace std; int n; int num[25]={0,1}; //输出数组 int prime[20]={2,3,5,7,11,13,17,19,23,29,31,37}; //先把1-20范围内任意两数相加可能得到的质数罗列出来,供之后查证用 int mark[21]; void print() { int i; for(i=1;i<n;i++) printf("%d ",num[i]); printf("%d\n",num[i]); } int isPrime(int x) //在所有可能的质数中找寻该数,方便查证该数是否为质数 { int i; for(i=0;i<12;i++) { if(x==prime[i]) return 1; } return 0; } int sear(int pre,int post,int flag) //深搜函数,pre为前一个节点,post为当前节点,flag是节点序号 { int i; if(!isPrime(pre+post)) //若两数之和不为质数,直接退出 return 0; num[flag]=post; //符合要求,将当前节点post保存到num数组对应位置中。 if(flag==n&&isPrime(post+1)) /*序号为n,即所有节点已经找全,且当前节点,即最末尾节点与头节点之和为质数,则打印该序列并返回1.*/ { print(); return 1; } mark[post]=0; //已使用过的数字标记为0 for(i=2;i<=n;i++) { if(mark[i]!=0&&sear(post,i,flag+1)) //接下来是递归搜索在当前序列下的后续可能序列,到达底部后再层层返回 break; } mark[post]=1; //已使用数字的标记恢复为1 return 0; } int main() { int i; int order=1; while(~scanf("%d",&n)) { printf("Case %d:\n",order++); for(i=1;i<=n;i++) //标记所有数字都可使用 mark[i]=1; if(n==1) printf("1\n"); for(i=2;i<=n;i++) //头元素固定为1,第二元素从2开始到n变化,深度优先搜索 sear(1,i,2); printf("\n"); //案例末尾换行 } return 0; }
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